Relative projectivity forces character vanishing off the controlling p-section

The assertion is immediate if $M=0$, so assume otherwise. Let $t$ be the $p$-part of $x$. Since $1\in Q$, the hypothesis implies $t\mathrm{\ne }1$; in particular $p$ divides $\mathrm{\mid }x\mathrm{\mid }$. Put $C=⟨x⟩\text{and}L=⟨x^p⟩\mathrm{.}$ Then $L⊴C$ and $[C:L]=p$. The subgroup generated by $t$ is the Sylow $p$-subgroup of the cyclic group $C$. If $R\le C$ does not contain $t$, then $R\le L$: indeed, any power $x^j\mathrm{\notin }L$ has $p\nmid j$ and generates the whole Sylow $p$-subgroup of $C$.

By F1, $M$ is a direct summand of $Y={\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^{H}M$. Restricting to $C$ and using F2 gives ${\mathrm{Res}}_C^{H}Y\cong {⨁}_{s\in C\mathrm{\backslash }H\mathrm{/}Q}{\mathrm{Ind}}_{R_s}^C{V}_s,R_s=C\cap sQ{s}^{-1},$ for the corresponding integral $R_s$-lattices $V_s$. If some $R_s$ contained $t$, then $t$ would belong to the $H$-conjugate $sQ{s}^{-1}$, contrary to the hypothesis. Thus step 1.1 gives $R_s\le L$ for every $s$. By induction transitivity, ${\mathrm{Ind}}_{R_s}^C{V}_s\cong {\mathrm{Ind}}_L^C\left({\mathrm{Ind}}_{R_s}^L{V}_s\right)\mathrm{.}$

Decompose each nonzero $\mathcal{O}L$-lattice ${\mathrm{Ind}}_{R_s}^L{V}_s$ into indecomposables $W_{s,j}$ by F3. Because $L⊴C$ has index $p$, F4 says that every ${\mathrm{Ind}}_L^C{W}_{s,j}$ is indecomposable. Hence step 2.1 is an indecomposable decomposition of ${\mathrm{Res}}_C^{H}Y$. Since ${\mathrm{Res}}_C^{H}M$ is a direct summand, Krull--Schmidt makes each of its indecomposable summands isomorphic to one of these induced lattices.

For any $\mathcal{O}L$-lattice $W$, scalar extension identifies $K{\otimes }_{\mathcal{O}}{\mathrm{Ind}}_L^{C}W$ with ${\mathrm{Ind}}_L^C(K{\otimes }_{\mathcal{O}}W)$. Its direct-sum decomposition over the $p$ cosets of $L$ in $C$ is cyclically permuted by $x$, because $x\mathrm{\notin }L$. Thus the matrix of $x$ has zero diagonal blocks, and its trace is zero. Applying this to every summand selected in step 3.1 and adding their traces yields $\theta (x)=\mathrm{tr}(x\mid K{\otimes }_{\mathcal{O}}M)=0.$ Algebraic closedness is used exactly through Green indecomposability in step 3.1; all selections and sums are finite.